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3 years ago

Solve the system of equationsy = 4x+1y=2+2x-2

Mathematics

1 answer:

Natalija [7]3 years ago

5 0

Answer:

x = -1/2 , y = -1

Step-by-step explanation:

Solve the following system:

{y = 4 x + 1 | (equation 1)

{y = 2 x | (equation 2)

Express the system in standard form:

{-(4 x) + y = 1 | (equation 1)

{-(2 x) + y = 0 | (equation 2)

Subtract 1/2 × (equation 1) from equation 2:

{-(4 x) + y = 1 | (equation 1)

{0 x+y/2 = (-1)/2 | (equation 2)

Multiply equation 2 by 2:

{-(4 x) + y = 1 | (equation 1)

{0 x+y = -1 | (equation 2)

Subtract equation 2 from equation 1:

{-(4 x)+0 y = 2 | (equation 1)

{0 x+y = -1 | (equation 2)

Divide equation 1 by -4:

{x+0 y = (-1)/2 | (equation 1)

{0 x+y = -1 | (equation 2)

Collect results:

Answer: {x = -1/2 , y = -1

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svet-max [94.6K]

Answer:

If k = −1 then the system has no solutions.

If k = 2 then the system has infinitely many solutions.

The system cannot have unique solution.

Step-by-step explanation:

We have the following system of equations

$x - 2y +3z = 2\\x + y + z = k\\2x - y + 4z = k^2$

The augmented matrix is

$\left[\begin{array}{cccc}1&-2&3&2\\1&1&1&k\\2&-1&4&k^2\end{array}\right]$

The reduction of this matrix to row-echelon form is outlined below.

$R_2\rightarrow R_2-R_1$

$\left[\begin{array}{cccc}1&-2&3&2\\0&3&-2&k-2\\2&-1&4&k^2\end{array}\right]$

$R_3\rightarrow R_3-2R_1$

$\left[\begin{array}{cccc}1&-2&3&2\\0&3&-2&k-2\\0&3&-2&k^2-4\end{array}\right]$

$R_3\rightarrow R_3-R_2$

$\left[\begin{array}{cccc}1&-2&3&2\\0&3&-2&k-2\\0&0&0&k^2-k-2\end{array}\right]$

The last row determines, if there are solutions or not. To be consistent, we must have k such that

$k^2-k-2=0$

$\left(k+1\right)\left(k-2\right)=0\\k=-1,\:k=2$

Case k = −1:

$\left[\begin{array}{ccc|c}1&-2&3&2\\0&3&-2&-1-2\\0&0&0&(-1)^2-(-1)-2\end{array}\right] \rightarrow \left[\begin{array}{ccc|c}1&-2&3&2\\0&3&-2&-3\\0&0&0&-2\end{array}\right]$

If k = −1 then the last equation becomes 0 = −2 which is impossible.Therefore, the system has no solutions.

Case k = 2:

$\left[\begin{array}{ccc|c}1&-2&3&2\\0&3&-2&2-2\\0&0&0&(2)^2-(2)-2\end{array}\right] \rightarrow \left[\begin{array}{ccc|c}1&-2&3&2\\0&3&-2&0\\0&0&0&0\end{array}\right]$

This gives the infinite many solution.

5 0

3 years ago

Solve the system of equationsy = 4x+1y=2+2x-2​

Solve the system of equationsy = 4x+1y=2+2x-2